Question 2.Q.64: In the circuit of Fig. 1, compute Vx . Demonstrate the rec......
In the circuit of Fig. 1, compute \overline{V}_x . Demonstrate the reciprocity theorem by interchanging the positions of the source and response.
Case i : To solve \overline{V}_x in the given circuit
Let us assume two node voltages \overline{V}_1 and \overline{V}_2 as shown in Fig. 2. Now, the response, \overline{V}_x=\overline{V}_2.
With reference to Fig. 2, the node basis matrix equation is,
\begin{bmatrix} {{\frac{1}{2+j4}+\frac{1}{j5}}}&{{-\frac{1}{j5}}}\\ {{-\frac{1}{j5}}}&{{\frac{1}{j5}+\frac{1}{8}+\frac{1}{4+j6}}} \end{bmatrix} \begin{bmatrix}\overline{ V} _{1} \\ \overline{ V}_{2} \end{bmatrix} =\begin{bmatrix} 10 \angle 60^\circ \\ 0 \end{bmatrix}
10\angle60^{\circ}=10\cos60^{\circ}+j10\sin60^{\circ}=5+j8.6603\,A\begin{bmatrix} {{0.1+j0.4}}&{{j0.2}}\\ {{j0.2}}&0.2019- j 0.3154 \end{bmatrix} \begin{bmatrix}\overline{ V} _{1} \\ \overline{ V}_{2} \end{bmatrix} =\;\left[\mathrm{5+j8.6603}\atop0\right] …………….(1)
\Delta ^{\prime}=\begin{vmatrix}{{0.1+j0.4}}&{{j0.2}}\\ {{j0.2}}&0.2019- j 0.3154 \end{vmatrix} =\left[(0.1-\mathrm{j0.4})\times(0.2019-\mathrm{j0.3154})\right]-(\mathrm{j0.2})^{2}=0.06597-\mathrm{j0.1123}
\Delta_2^{\prime}=\left|\begin{array}{rr}0.1-j 0.4 & 5+j 8.6603 \\j 0.2 & 0\end{array}\right|\begin{aligned}& =0-[j 0.2 \times(5+j 8.6603)] \\& =1.73206- j\end{aligned}
∴ The response,\overline{{{V}}}_{x}\,=\,\overline{V}_{2}\,=\,\frac{\Delta^{'}_2}{\Delta^{'}}\,=\,\frac{1.73206-\mathrm{j{}}}{-0.06597-\mathrm{j0.1123}}
=\;-0.1158+j15.3555\;V
=\,1\,5.3559∠90.4^{\circ}\,V
Case ii : To demonstrate the reciprocity theorem by interchanging the positions of source and response
Let us interchange the positions of source and response as shown in Fig. 3. Let us assume node voltages \overline{V}_a and \overline{V}_b as shown in Fig. 3.
Now, the response ,\overline{{V}}_{x}\;=\;\overline{{V}}_{a} With reference to Fig. 3, the node basis matrix equation is,
\begin{bmatrix} {{\frac{1}{2+j4}+\frac{1}{j5}}}&{{-\frac{1}{j5}}}\\ {{-\frac{1}{j5}}}&{{\frac{1}{j5}+\frac{1}{8}+\frac{1}{4+j6}}} \end{bmatrix} \begin{bmatrix}\overline{ V} _{a} \\ \overline{ V}_{b} \end{bmatrix} =\begin{bmatrix} 0 \\10 \angle 60^\circ \end{bmatrix}
\begin{bmatrix} {{0.1-j0.4}}&{{j0.2}}\\ {{j0.2}}&0.2019- j 0.3154 \end{bmatrix} \begin{bmatrix}\overline{ V} _{a} \\ \overline{ V}_{b} \end{bmatrix} =\;\left[0\atop\mathrm{5+j8.6603}\right] ……………(2)
On comparing equations (1) and (2), we can say that the value of Δ^{'} remains the same.
∴ \Delta^{\prime}\,=\,-0.06597-\mathrm{j0.1123}=
\Delta ^{\prime}_a=\begin{vmatrix}{{0}}&{{j0.2}}\\ {{5+j 8.6603}}&0.2019- j 0.3154 \end{vmatrix} =0-\left[(5+\mathrm{j}8.6603)\times\mathrm{j}0.2\right]=1 .73206 − j∴ The response, \overline{{{V}}}_{x}\,=\,\overline{V}_{a}\,=\,\frac{\Delta^{'}_a}{\Delta^{'}}\,=\,\frac{1.73206-\mathrm{j{}}}{-0.06597-\mathrm{j0.1123}}
=\;-0.1158+j15.3555\;V
=\,15.3559∠90.4^{\circ}\,V
It is observed that the response remains the same after interchanging the positions of source and response, which demonstrates the validity of the reciprocity theorem.
